已知丨a丨=2√2,丨b丨=3,a,b的夹角为π/4,若向量AB=5a+2b,向量AC=a-3b,且D为BC重点,则向量AD的长度为

问题描述:

已知丨a丨=2√2,丨b丨=3,a,b的夹角为π/4,若向量AB=5a+2b,向量AC=a-3b,且D为BC重点,则向量AD的长度为

∵a.b=丨a丨丨b丨cos(π/4)=6 AD=1/2(AB+AC)=1/2(5a+2b+a-3b)=3a-1/2b AD^2=(3a-1/2b)^2=9a^2-3a.b+1/4^b2 =9*(2√2)^2-3*6+(1/4)*9 =225/4 ∴ |AD|=15/2